Chern classes of tautological bundle over the Grassmannian G(2,4)












8












$begingroup$


I've the following problem:



I know how to calculate Chern classes of the tautological bundle over the Grassmannian $G=G(2,4)$ using the Schubert calculus. If I am right, the Chern character should be
$$1-sigma_1+sigma_{1,1}$$
where $sigma_{i_1,i_2}$ indicates the Schubert cycle corresponding to the Schubert variety $Sigma_{i_1,i_2}={Lambdain Gmiddim(V_{2-i_j+j}capLambda)geq j;forall j}$ (here ${V_j}subset V$ is a flag in the 4-dimensional vector space $V$). Now these classes are element in the Chow ring, but I want to work with classes in the integral cohomology group. How can I do this translation?



Thank you!






algebraic-geometry schubert-calculus







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$endgroup$



migrated from mathoverflow.net Jul 15 '14 at 22:07


This question came from our site for professional mathematicians.














  • 3




    $begingroup$
    The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
    $endgroup$
    – abx
    Jul 15 '14 at 14:28










  • $begingroup$
    I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
    $endgroup$
    – Cla
    Jul 17 '14 at 9:25
















8












$begingroup$


I've the following problem:



I know how to calculate Chern classes of the tautological bundle over the Grassmannian $G=G(2,4)$ using the Schubert calculus. If I am right, the Chern character should be
$$1-sigma_1+sigma_{1,1}$$
where $sigma_{i_1,i_2}$ indicates the Schubert cycle corresponding to the Schubert variety $Sigma_{i_1,i_2}={Lambdain Gmiddim(V_{2-i_j+j}capLambda)geq j;forall j}$ (here ${V_j}subset V$ is a flag in the 4-dimensional vector space $V$). Now these classes are element in the Chow ring, but I want to work with classes in the integral cohomology group. How can I do this translation?



Thank you!






algebraic-geometry schubert-calculus







share|cite|improve this question











$endgroup$



migrated from mathoverflow.net Jul 15 '14 at 22:07


This question came from our site for professional mathematicians.














  • 3




    $begingroup$
    The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
    $endgroup$
    – abx
    Jul 15 '14 at 14:28










  • $begingroup$
    I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
    $endgroup$
    – Cla
    Jul 17 '14 at 9:25














8












8








8


2



$begingroup$


I've the following problem:



I know how to calculate Chern classes of the tautological bundle over the Grassmannian $G=G(2,4)$ using the Schubert calculus. If I am right, the Chern character should be
$$1-sigma_1+sigma_{1,1}$$
where $sigma_{i_1,i_2}$ indicates the Schubert cycle corresponding to the Schubert variety $Sigma_{i_1,i_2}={Lambdain Gmiddim(V_{2-i_j+j}capLambda)geq j;forall j}$ (here ${V_j}subset V$ is a flag in the 4-dimensional vector space $V$). Now these classes are element in the Chow ring, but I want to work with classes in the integral cohomology group. How can I do this translation?



Thank you!






algebraic-geometry schubert-calculus







share|cite|improve this question











$endgroup$




I've the following problem:



I know how to calculate Chern classes of the tautological bundle over the Grassmannian $G=G(2,4)$ using the Schubert calculus. If I am right, the Chern character should be
$$1-sigma_1+sigma_{1,1}$$
where $sigma_{i_1,i_2}$ indicates the Schubert cycle corresponding to the Schubert variety $Sigma_{i_1,i_2}={Lambdain Gmiddim(V_{2-i_j+j}capLambda)geq j;forall j}$ (here ${V_j}subset V$ is a flag in the 4-dimensional vector space $V$). Now these classes are element in the Chow ring, but I want to work with classes in the integral cohomology group. How can I do this translation?



Thank you!






algebraic-geometry schubert-calculus




algebraic-geometry schubert-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 0:49









Matt Samuel

38.7k63769




38.7k63769










asked Jul 15 '14 at 14:07









ClaCla

638411




638411




migrated from mathoverflow.net Jul 15 '14 at 22:07


This question came from our site for professional mathematicians.









migrated from mathoverflow.net Jul 15 '14 at 22:07


This question came from our site for professional mathematicians.










  • 3




    $begingroup$
    The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
    $endgroup$
    – abx
    Jul 15 '14 at 14:28










  • $begingroup$
    I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
    $endgroup$
    – Cla
    Jul 17 '14 at 9:25














  • 3




    $begingroup$
    The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
    $endgroup$
    – abx
    Jul 15 '14 at 14:28










  • $begingroup$
    I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
    $endgroup$
    – Cla
    Jul 17 '14 at 9:25








3




3




$begingroup$
The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
$endgroup$
– abx
Jul 15 '14 at 14:28




$begingroup$
The natural map $CH(G)rightarrow H^*(G,mathbb{Z})$ is a ring isomorphism.
$endgroup$
– abx
Jul 15 '14 at 14:28












$begingroup$
I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
$endgroup$
– Cla
Jul 17 '14 at 9:25




$begingroup$
I know that a such map exists, but I don't how it works. So in my specific example, what is the image of $-sigma_1$? And that of $sigma_{1,1}$? I never did such a calculation..
$endgroup$
– Cla
Jul 17 '14 at 9:25










1 Answer
1






active

oldest

votes


















10












$begingroup$

The cohomology ring of $G(k,n)$ is isomorphic to
$$frac{mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$
where $T$ and $Q$ are respectively the tautological and the quotient bundle.
The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas:




  • $c_i(T) = (−1)^isigma_{1,...,1}$,

  • $c_i(Q) = sigma_i$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
    $endgroup$
    – Jason Starr
    Jul 15 '14 at 17:15










  • $begingroup$
    You are right. I corrected it.
    $endgroup$
    – A. M. Flat_Line
    Jul 15 '14 at 19:16











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1 Answer
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1 Answer
1






active

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active

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active

oldest

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10












$begingroup$

The cohomology ring of $G(k,n)$ is isomorphic to
$$frac{mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$
where $T$ and $Q$ are respectively the tautological and the quotient bundle.
The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas:




  • $c_i(T) = (−1)^isigma_{1,...,1}$,

  • $c_i(Q) = sigma_i$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
    $endgroup$
    – Jason Starr
    Jul 15 '14 at 17:15










  • $begingroup$
    You are right. I corrected it.
    $endgroup$
    – A. M. Flat_Line
    Jul 15 '14 at 19:16
















10












$begingroup$

The cohomology ring of $G(k,n)$ is isomorphic to
$$frac{mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$
where $T$ and $Q$ are respectively the tautological and the quotient bundle.
The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas:




  • $c_i(T) = (−1)^isigma_{1,...,1}$,

  • $c_i(Q) = sigma_i$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
    $endgroup$
    – Jason Starr
    Jul 15 '14 at 17:15










  • $begingroup$
    You are right. I corrected it.
    $endgroup$
    – A. M. Flat_Line
    Jul 15 '14 at 19:16














10












10








10





$begingroup$

The cohomology ring of $G(k,n)$ is isomorphic to
$$frac{mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$
where $T$ and $Q$ are respectively the tautological and the quotient bundle.
The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas:




  • $c_i(T) = (−1)^isigma_{1,...,1}$,

  • $c_i(Q) = sigma_i$.






share|cite|improve this answer









$endgroup$



The cohomology ring of $G(k,n)$ is isomorphic to
$$frac{mathbb{Z}[c_1(T),...,c_k(T),c_1(Q),..., c_{n−k} (Q)]}{(c(T)c(Q) = 1)},$$
where $T$ and $Q$ are respectively the tautological and the quotient bundle.
The chern classes of the tautological and the quotient bundle are given in terms of Schubert cycles by the following formulas:




  • $c_i(T) = (−1)^isigma_{1,...,1}$,

  • $c_i(Q) = sigma_i$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jul 15 '14 at 15:54









A. M. Flat_LineA. M. Flat_Line

37916




37916








  • 1




    $begingroup$
    In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
    $endgroup$
    – Jason Starr
    Jul 15 '14 at 17:15










  • $begingroup$
    You are right. I corrected it.
    $endgroup$
    – A. M. Flat_Line
    Jul 15 '14 at 19:16














  • 1




    $begingroup$
    In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
    $endgroup$
    – Jason Starr
    Jul 15 '14 at 17:15










  • $begingroup$
    You are right. I corrected it.
    $endgroup$
    – A. M. Flat_Line
    Jul 15 '14 at 19:16








1




1




$begingroup$
In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
$endgroup$
– Jason Starr
Jul 15 '14 at 17:15




$begingroup$
In the displayed formula, you should have a $mathbb{Z}$ where you have a "coefficient field" $k$.
$endgroup$
– Jason Starr
Jul 15 '14 at 17:15












$begingroup$
You are right. I corrected it.
$endgroup$
– A. M. Flat_Line
Jul 15 '14 at 19:16




$begingroup$
You are right. I corrected it.
$endgroup$
– A. M. Flat_Line
Jul 15 '14 at 19:16


















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